Chapter 6-the rest

Question 1

Casting

Answer 1

• In the casting process, metals are first melted in a furnace.
• Alloying elements are then added and thoroughly mixed
  (magnesium is added to aluminum to produce a stronger and
  lighter alloy).
• Oxide impurities and unwanted gasses are removed.
• The melt is then poured into a mold and chilled to solidify

Question 2

recrystallization temp

Answer 2

Used in hot rolling where materials are heated to a high temp

Question 3

% cold work

Answer 3

= [(initial metal thickness- final thickness)/ initial thickness] x 100

Question 4

extrusion

Answer 4

metal workpiece (billet) is placed under high
pressure using a ram and forced through opening in a die

Question 4

indirect extrusion

Answer 4

experiences lower friction on the billet and
therefore needs less power however has a limit on the applied
load

Question 5

direct extrusion

Answer 5

produces products such as cylindrical rods, tubes,
and with some metals more irregular shapes

Question 6

forging

Answer 6

metal is hammered into a desired shape

Question 7

open die

Answer 7

Dies are either flat or
simple in geometry

Question 8

closed die

Answer 8

Dies have upper
and lower impression – more complex

Question 9

wire drawing

Answer 9

the starting rod or wire is drawn through
several drawing dies to reduce diameter

Question 10

cold work of wire drawing

Answer 10

= (Change in cross-sectional area/ original area) x 100

Question 11

elastic deformation

Answer 11

Atoms elongate resulting in overall elongation of the specimen. But
return to their original dimensions after tensile force is removed,
resulting in recoverable deformation

Question 12

plastic deformation

Answer 12

Atoms break bonds and slip on each other.
Once slip occurs, atoms can not return to their original positions, resulting in permanent deformation.

Question 13

normal stress σ

Answer 13

=(F -> avg uniaxial tensile force)/ (Ao -> original cross-sectional area) (lb/in^2 or N/m^2)

Question 14

Normal strain ε

Answer 14

=Change in length/ original length

Question 15

shear stress τ (parallel to surface)

Answer 15

=(S -> shear force)/ (A -> area of application)

Question 16

shear stress γ (change in shape)

Answer 16

= amount of shear displacement a/ distance h over which stress acts = tan θ

Question 17

linear elastic range

Answer 17

linear portion of the stress strain curve

Question 18

yield point

Answer 18

The point on the curve when the elastic range ends

Question 19

ultimate tensile strength σu

Answer 19

the largest stress the
specimen can take before fracture

Question 20

Fracture point

Answer 20

The point on the curve where the specimen fails

Question 21

Modulus of elasticity (E)

Answer 21

Stress and strain are linearly related in the elastic region. (Hooke’s law)

= stress/ strain

Question 22

poisson’s ratio is always positive

Answer 22

= -[(w-wo)/ wo] / [(l-lo)/lo]

Question 23

Yield strength

Answer 23

the stress at which
a material begins to experience
permanent or plastic deformation

Question 24

% elongation fracture

Answer 24

a measure of ductility of a material = (final length- initial length)/ initial length

Question 25

modulus of resilience Ur

Answer 25

measure of the amount of energy needed to cause yield in the material =1/2 (σy-Ey)

Question 26

toughness

Answer 26

amount of energy required to fracture the material. It is a measure of combination of strength and ductility

Question 27

true stress and true strain

Answer 27

are determined based on instantaneous cross-sectional area and length of the specimen

Question 28

hardness

Answer 28

a measure of the resistance of a metal to localized plastic deformation or indentation

Question 29

slip bands

Answer 29

Plastic deformation in a single crystal results in formation of bands

Question 30

slip

Answer 30

caused by application of shear stress on the grain. The mechanism by which slip occurs is by formation of dislocations and their movement

Question 31

unit step of slip

Answer 31

created Once the dislocation reaches the end of the grain

Question 32

resolved shear stress (shimid's law)

Answer 32

Tr=σ(cosλ cosθ)

Question 33

critical resolved stress

Answer 33

the shear stress required to cause slip in a single crystal

Question 34

twin

Answer 34

part of atomic lattice is deformed and forms mirror image of lattice next to it

Question 35

hall-petch equation

Answer 35

=σy = σ0 + k/d^1/2 σy= yield strength σ0 and k are constants d is avg grain size diameter

Question 36

solid solutions

Answer 36

addition of one or more impurity (solute) to the host metal can increase the strength of metals

Question 37

annealing

Answer 37

heat treatment process applied to a strain- hardened or cold-worked metal to remove residual stresses, grow new equiaxed grains and soften the metal three stages of the annealing process are: recovery, recrystallization and grain growth

Question 38

polgonization

Answer 38

Dislocations rearrange themselves to form lower energy configurations

Question 39

composite

Answer 39

combination of two or more individual materials designed to obtain more individual properties

Question 40

composite phase types

Answer 40

-- Matrix - is continuous -- Dispersed - is discontinuous and surrounded by matrix

Question 41

Composite Types:

Answer 41

Particle reinforced Fiber reinforced structural nano

Question 42

Particle reinforced

Answer 42

-- Types: large-particle and dispersion-strengthened -- Properties are isotropic

Question 43

fiber reinforced can be continuous (aligned) and discontinuous (aligned or random)

Answer 43

* Fibers very strong in tension – Provide significant strength improvement to the composite – Properties can be isotropic or anisotropic

Question 44

structural

Answer 44

* Laminates - -- stacked and bonded fiber-reinforced sheets * Sandwich panels -- honeycomb core between two facing sheets

Question 45

pultrusion

Answer 45

* Continuous fibers pulled through resin tank to impregnate fibers with thermosetting resin * Impregnated fibers pass through steel die that preforms to the desired shape * Preformed stock passes through a curing die that is – precision machined to impart final shape – heated to initiate curing of the resin matrix

Question 46

filament windind

Answer 46

– Continuous reinforcing fibers are accurately positioned in a predetermined pattern to form a hollow (usually cylindrical) shape – Fibers are fed through a resin bath to impregnate with thermosetting resin – Impregnated fibers are continuously wound (typically automatically) onto a mandrel – After appropriate number of layers added, curing is carried out either in an oven or at room temperature – The mandrel is removed to give the final product

Question 47

Ceramic Materials

Answer 47

Clay glass refractories carbon abrasives cements

Question 48

Die blanks

Answer 48

-- Need wear resistant properties! * Die surface: -- 4 μm polycrystalline diamond particles that are sintered onto a cemented tungsten carbide substrate

Question 49

refractories

Answer 49

* Materials to be used at high temperatures (e.g., in high temperature furnaces)

Question 50

Ceramics

Answer 50

the bonding between the atoms is generally covalent or ionic, and they are much stronger than the metallic bonds.  Hardness, thermal and electrical resistance are significantly higher than in metals.  Ceramics are available in single-crystal and polycrystalline form.  Finer grain size ceramics have higher strength and toughness.

Question 51

phase change

Answer 51

the structure of a material changes from one form to another.

Question 52

phase

Answer 52

A region in a material that has a uniform structure, properties, and composition and maintains a distinct boundary with other unlike phases. Ex: ice, water, vapor

Question 53

equilibrium

Answer 53

The state of a system when all forces and energies are balanced resulting in a stable system with no tendency to change with time

Question 54

phase diagram

Answer 54

Graphical representations of phases present in a material system at different temperatures, pressures and compositions. Are developed based on assumption of equilibrium conditions resulting from slow cooling (equilibrium is approached but never fully maintained)

Question 55

triple point

Answer 55

point at which the 3 stages of matter coexist

Question 56

The number of phases that can co-exist in equilibrium in a given system. * P+F = C+2

Answer 56

P = number of phases that coexist in a system C = Number of components F = Degrees of freedom

Question 57

cooling curve

Answer 57

Temperature of molten metal is recorded versus time as it cools to room temperature, Every time there is a phase change, there is a change in slope. For a pure metal, the cooling curve shows a flat region (plateau) at a specific temperature where liquid transforms to solid

Question 58

In plateau region:

Answer 58

there is a mixture of two phases in equilibrium.

Question 59

In a phase diagram:

Answer 59

Created from a series of cooling curves. * Above the “start” line, the alloy is all liquid. * Below the “finish” line, the alloy is all solid. In the region between start and finish, a mixture of solid and liquid exists

Question 60

liquidius

Answer 60

The line that indicates start of the solidification process

Question 61

solidus

Answer 61

The line that indicates completion of the solidification process

Question 62

binary isomorphous system

Answer 62

a system where the two components are completely soluble in each other in both liquid and solid states (they do not from a third phase)

Question 63

Tie line

Answer 63

In the mixture region (L + α), the composition of both liquid (L) and solid (α) phases at any temperature can be determined by drawing a tie line

Question 64

Wo

Answer 64

Overall alloy composition

Question 65

Wl

Answer 65

The composition at point L is the composition of liquid, in the mixture.

Question 66

Ws

Answer 66

The composition at point S is the composition of solid, in the mixture.

Question 67

The amount of each phase can be determined through application of the Lever Rule:

Answer 67

Wt fraction of solid phase = Xs= (Wo-Wl)/(Ws-Wl) Wt fraction in liquid phase= Xl= (Ws-Wo)/(Ws-Wl)